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Find the value of ∫ from 0 to 1 of (x^3 - 3x^2 + 3x) dx.
Find the value of ∫ from 0 to 1 of (x^3 - 3x^2 + 3x) dx.
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Q1
Find the value of ∫ from 0 to 1 of (x^3 - 3x^2 + 3x) dx.
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The integral evaluates to [x^4/4 - x^3 + (3/2)x^2] from 0 to 1 = (1/4 - 1 + 3/2) = 1/4.
Questions & Step-by-step Solutions
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Q
Q: Find the value of ∫ from 0 to 1 of (x^3 - 3x^2 + 3x) dx.
Solution:
The integral evaluates to [x^4/4 - x^3 + (3/2)x^2] from 0 to 1 = (1/4 - 1 + 3/2) = 1/4.
Steps: 7
Show Steps
Step 1: Identify the integral you need to solve: ∫ from 0 to 1 of (x^3 - 3x^2 + 3x) dx.
Step 2: Find the antiderivative of the function (x^3 - 3x^2 + 3x).
Step 3: The antiderivative is calculated as follows: For x^3, the antiderivative is x^4/4; for -3x^2, it is -x^3; and for 3x, it is (3/2)x^2.
Step 4: Combine the antiderivatives: The complete antiderivative is (x^4/4 - x^3 + (3/2)x^2).
Step 5: Evaluate the antiderivative from 0 to 1: Substitute 1 into the antiderivative: (1^4/4 - 1^3 + (3/2)(1^2)) = (1/4 - 1 + 3/2).
Step 6: Simplify the expression: 1/4 - 1 + 3/2 = 1/4 - 4/4 + 6/4 = (1 - 4 + 6)/4 = 3/4.
Step 7: The final answer is 3/4.
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